CODEWITHSUNDEEP
node.jsmongodbmongoose
Kuba T
Kuba T

Reputation: 3083

Mongoose - finding subdocuments by criteria

I've just got stuck with this problem. I've got two Mongoose schemas:

var childrenSchema = mongoose.Schema({
    name: {
        type: String
    },
    age: {
        type: Number,
        min: 0
    }
});

var parentSchema = mongoose.Schema({
    name : {
        type: String
    },
    children: [childrenSchema]
});

Question is, how to fetch all subdocuments (in this case, childrenSchema objects) from every parent document? Let's suppose I have some data:

var parents = [
    { name: "John Smith",
    children: [
        { name: "Peter", age: 2 }, { name: "Margaret", age: 20 }
    ]},
    { name: "Another Smith",
    children: [
        { name: "Martha", age: 10 }, { name: "John", age: 22 }
    ]}
];

I would like to retrieve - in a single query - all children older than 18. Is it possible? Every answer will be appreciated, thanks!

Upvotes: 38

Views: 52507

Answers (3)

cjwiseman11
cjwiseman11

Reputation: 74

A. Jesse Jiryu Davis's response works like a charm, however for later versions of Mongoose (Mongoose 5.x) we get the error:

Mongoose 5.x disallows passing a spread of operators to Model.aggregate(). Instead of Model.aggregate({ $match }, { $skip }), do Model.aggregate([{ $match }, { $skip }])

So the code would simply now be:

> db.parents.aggregate([{
    $match: {'children.age': {$gte: 18}}
}, {
    $unwind: '$children'
}, {
    $match: {'children.age': {$gte: 18}}
}, {
    $project: {
        name: '$children.name',
        age:'$children.age'
    }
}])
{
    "result" : [
        {
            "_id" : ObjectId("51a7bf04dacca8ba98434eb5"),
            "name" : "Margaret",
            "age" : 20
        },
        {
            "_id" : ObjectId("51a7bf04dacca8ba98434eb6"),
            "name" : "John",
            "age" : 22
        }
    ],
    "ok" : 1
}

(note the array brackets around the queries)

Hope this helps someone!

Upvotes: 1

AbdelHady
AbdelHady

Reputation: 9712

In Mongoose, you can also use the elegant .populate() function like this:

parents
.find({})
.populate({
  path: 'children',
  match: { age: { $gte: 18 }},
  select: 'name age -_id'
})
.exec()

Upvotes: 23

A. Jesse Jiryu Davis
A. Jesse Jiryu Davis

Reputation: 24017

You can use $elemMatch as a query-projection operator in the most recent MongoDB versions. From the mongo shell:

db.parents.find(
    {'children.age': {$gte: 18}},
    {children:{$elemMatch:{age: {$gte: 18}}}})

This filters younger children's documents out of the children array:

{ "_id" : ..., "children" : [ { "name" : "Margaret", "age" : 20 } ] }
{ "_id" : ..., "children" : [ { "name" : "John", "age" : 22 } ] }

As you can see, children are still grouped inside their parent documents. MongoDB queries return documents from collections. You can use the aggregation framework's $unwind method to split them into separate documents:

> db.parents.aggregate({
    $match: {'children.age': {$gte: 18}}
}, {
    $unwind: '$children'
}, {
    $match: {'children.age': {$gte: 18}}
}, {
    $project: {
        name: '$children.name',
        age:'$children.age'
    }
})
{
    "result" : [
        {
            "_id" : ObjectId("51a7bf04dacca8ba98434eb5"),
            "name" : "Margaret",
            "age" : 20
        },
        {
            "_id" : ObjectId("51a7bf04dacca8ba98434eb6"),
            "name" : "John",
            "age" : 22
        }
    ],
    "ok" : 1
}

I repeat the $match clause for performance: the first time through it eliminates parents with no children at least 18 years old, so the $unwind only considers useful documents. The second $match removes $unwind output that doesn't match, and the $project hoists children's info from subdocuments to the top level.

Upvotes: 61

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